Skip to content
Worked Examples · Example 26

Q.A man saves ₹500 in the first month and in successive months he saves twice as much as in the previous month. This process continued for 6 months. From the seventh month and onwards he is able to save ₹500 less than the previous month. Find his total savings for the year.

Yanam CbseNCERTSubjective· 3mImportance★★★★★est
71% · 75/106 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Split the year into a 6-term G.P. (doubling savings) and a 6-term A.P. (decreasing savings), and add the two partial sums.

[!FORMULA] G.P. sum: Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}. A.P. sum: Sn=n2[2a+(n−1)d]S_n=\dfrac{n}{2}\left[2a+(n-1)d\right], where aa = first term, rr/dd = common ratio/difference, nn = number of terms.

  1. Months 1–6 (G.P.): a=₹500a=\text{₹}500, r=2r=2 (savings double each month), n=6n=6.
  2. S6=500(26−1)2−1=500(63)=₹31500S_{6}=\dfrac{500(2^{6}-1)}{2-1}=500(63)=\text{₹}31500.
  3. The 6th-month saving is T6=500×25=₹16000T_6=500\times2^{5}=\text{₹}16000.
  4. Months 7–12 (A.P.): from the 7th month, savings drop by ₹500 from the previous month, so the 7th-month saving is 16000−500=₹1550016000-500=\text{₹}15500. This is the A.P. first term: a′=15500a'=15500, d=−500d=-500, n=6n=6 (months 7 through 12). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.